A075485 Length of iteration list when Collatz-function is iterated with initial value 2^n - 1.

1, 8, 17, 18, 107, 108, 47, 48, 62, 63, 157, 158, 159, 160, 130, 131, 225, 226, 178, 179, 304, 305, 474, 475, 445, 446, 385, 386, 449, 450, 451, 452, 528, 529, 530, 531, 532, 533, 534, 535, 536, 537, 587, 588, 589, 590, 591, 592, 593, 594, 595, 596, 853, 854

Offset: 1

Author

Labos Elemer, Sep 26 2002

Keywords

• nonn

Comments

Somewhat surprisingly, these iterations take almost twice as long as the iterations for 2^n + 1. See A075486. - T. D. Noe, Jan 17 2013

Examples

			n=4, 2^n - 1 = 15, list = {15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1}, so a(4) = 18.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006370, A008908, A074472, A075484, A075486, A075487, A075488.

Programs

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Table[Length[Collatz[2^n - 1]], {n, 100}] (* T. D. Noe, Jan 17 2013 *)

Formula

a(n) = A008908(2^n-1).